// 例题10  离海最远的点（Most Distant Point from the Sea, Tokyo 2007, UVa1396）
// 刘汝佳
#include <cmath>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

struct Point {
  double x, y;
  Point(double x = 0, double y = 0): x(x), y(y) { }
};

typedef Point Vector;
Vector operator + (const Vector& A, const Vector& B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (const Point& A, const Point& B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (const Vector& A, double p) { return Vector(A.x * p, A.y * p); }
double Dot(const Vector& A, const Vector& B) { return A.x * B.x + A.y * B.y; }
double Cross(const Vector& A, const Vector& B) { return A.x * B.y - A.y * B.x; }
double Length(const Vector& A) { return sqrt(Dot(A, A)); }
Vector Normal(const Vector& A) { double L = Length(A); return Vector(-A.y / L, A.x / L); }

double PolygonArea(vector<Point> p) {
  int n = p.size();
  double area = 0;
  for (int i = 1; i < n - 1; i++)
    area += Cross(p[i] - p[0], p[i + 1] - p[0]);
  return area / 2;
}

// 有向直线。它的左边就是对应的半平面
struct Line {
  Point P;    // 直线上任意一点
  Vector v;   // 方向向量
  double ang; // 极角，即从x正半轴旋转到向量v所需要的角（弧度）
  Line() {}
  Line(const Point& P, const Vector& v): P(P), v(v) { ang = atan2(v.y, v.x); }
  bool operator < (const Line& L) const {
    return ang < L.ang;
  }
};

// 点p在有向直线L的左边（线上不算）
bool OnLeft(const Line& L, const Point& p) {
  return Cross(L.v, p - L.P) > 0;
}

// 二直线交点，假定交点惟一存在
Point GetLineIntersection(const Line& a, const Line& b) {
  Vector u = a.P - b.P;
  double t = Cross(b.v, u) / Cross(a.v, b.v);
  return a.P + a.v * t;
}

const double INF = 1e8, eps = 1e-6;

// 半平面交主过程
vector<Point> HalfplaneIntersection(vector<Line> L) {
  int n = L.size();
  sort(L.begin(), L.end()); // 按极角排序

  int first, last;         // 双端队列的第一个元素和最后一个元素的下标
  vector<Point> p(n);      // p[i]为q[i]和q[i+1]的交点
  vector<Line> q(n);       // 双端队列
  vector<Point> ans;       // 结果

  q[first = last = 0] = L[0]; // 双端队列初始化为只有一个半平面L[0]
  for (int i = 1; i < n; i++) {
    while (first < last && !OnLeft(L[i], p[last - 1])) last--;
    while (first < last && !OnLeft(L[i], p[first])) first++;
    q[++last] = L[i];
    if (fabs(Cross(q[last].v, q[last - 1].v)) < eps) { // 两向量平行且同向，取内侧的一个
      last--;
      if (OnLeft(q[last], L[i].P)) q[last] = L[i];
    }
    if (first < last) p[last - 1] = GetLineIntersection(q[last - 1], q[last]);
  }
  while (first < last && !OnLeft(q[first], p[last - 1])) last--; // 删除无用平面
  if (last - first <= 1) return ans; // 空集
  p[last] = GetLineIntersection(q[last], q[first]); // 计算首尾两个半平面的交点

  // 从deque复制到输出中
  for (int i = first; i <= last; i++) ans.push_back(p[i]);
  return ans;
}

int main() {
  for (int n, x, y; scanf("%d", &n) == 1 && n;) {
    vector<Vector> p, v, normal;
    for (int i = 0; i < n; i++) scanf("%d%d", &x, &y), p.push_back(Point(x, y));
    if (PolygonArea(p) < 0) reverse(p.begin(), p.end());

    for (int i = 0; i < n; i++)
      v.push_back(p[(i + 1) % n] - p[i]), normal.push_back(Normal(v[i]));
    double left = 0, right = 20000;
    while (right - left > eps) {
      vector<Line> L;
      double mid = left + (right - left) / 2;
      for (int i = 0; i < n; i++) L.push_back(Line(p[i] + normal[i]*mid, v[i]));
      vector<Point> poly = HalfplaneIntersection(L);
      if (poly.empty()) right = mid; else left = mid;
    }
    printf("%.6lf\n", left);
  }
  return 0;
}
// 25877776	1396	Most Distant Point from the Sea	Accepted	C++	0.000	2020-12-23 06:47:13